Young’s Double Slit Experiment

Apparatusoptics benchlaserslit filmscreenwhite paper and tapepencilmetric rulerOcean Optics spectrometer and fiber optics cable

GoalIn this experiment, you will set up a Double Slit apparatus and will measure the distance between brightfringes of the interference pattern. You will repeat the experiment for different pairs of slits to see how theinterference pattern depends on slit separation. You will use a curve fit to measure the wavelength of thelight.

IntroductionAs a plane wave passes through a slit, it emerges as circular waves as if from a point source of light. If lightfrom a laser (which is a plane wave) passes through two slits, then each slit acts as a point source of light.Because the incident wave is a plane wave and is monochromatic (i.e. one color, one frequency), then thewaves from each slit will be in phase (meaning a crest will emanate from each slit at the same time) and willhave the same frequency and amplitude.

The waves from each source will interfere. At some points in space, total constructive interference occursand a detector at this location will measure a wave with twice the amplitude, 2A, as the wave from eachsource. At other points in space total destructive interference occurs and a detector at this location willmeasure a wave with zero amplitude (meaning no wave at all). At the majority of points in space, interferenceresults in a wave that has an amplitude between 0 and 2A–meaning that it is not total constructive ordestructive interference. You an identify such points in Figure 1.

Figure 1: Constructive and destructive interference due to two sources.

At some points in space around the two sources, total constructive interference will occur; at these pointsthe path difference (∆L) from the two sources is:

∆L = |L1 − L2| = mλ m = 0, 1, 2, . . . total constructive interference

1

where L1 and L2 are the distances from the two sources to a point in space. For all points where m = 0,then L1 = L2 and total constructive interference occurs. One such location is shown in Figure 2.

Figure 2: Total constructive interference where m = 0.

You can identify a line of points where m = 0. Sketch a line on Figure 2 that connects all pointswhere m = 0 and total constructive interference occurs.

For all points where m = 1, then |L1−L2| = 1λ and total constructive interference occurs. Two differentlocations where m = 1 are shown in Figure 3.

Figure 3: Total constructive interference where m = 1.

You can identify two separate lines of points where m = 1. Sketch both lines on Figure 3 thatconnects all points where m = 1 and total constructive interference occurs.

Note that the points around the lines of total constructive interference are also bright and are a resultof constructive interference, but are not as bright as the points where total constructive interference occurs.These regions are called “fringes” and it is in the center of the fringes where total constructive interferenceoccurs. In the example above, there is a total of 5 fringes shown.

As you can imagine, the pattern of points where total constructive interference occurs is related to thewavelength of the light and the distance between the sources. To explore this, you will first use a simulation.

2

Experiment 1 – Exploring 2-source interference patterns with a simulation

Procedure1. Go to the URL:

http://physics.highpoint.edu/∼atitus/physlets/physlet.php?filename=optics_ripple-tank.html

2. A Java applet called Ripple Tank should load and you should see Figure 4. If the browser says“Missing Plugin" and you do not see Figure 4, then you may need to download and install Java fromwww.java.com/getjava/.

Figure 4: An applet to explore 2-D superposition of waves.

3. Click the link to d = λ and wait about 15 seconds for all of the frames to be calculated. Then it willrun and show the waves in motion. You can use the control buttons below the animation to stop, play,etc. In this case, the variable d is the distance between the sources. (Do not confuse this with L1 andL2 which were the distances from the sources to a point in space.)

How many interference “fringes” do you see in the animation window? (Don’t count any fringesbelow the sources in the −y direction.)

4. Now click the link to d = 2λ.

How many interference “fringes” do you see in the animation window? (Don’t count any fringesbelow the sources in the −y direction.)

5. Now click the link to d = 3λ.

How many interference “fringes” do you see in the animation window? (Don’t count any fringesbelow the sources in the −y direction.)

6. Now click the link to d = 4λ.

How many interference “fringes” do you see in the animation window? (Don’t count any fringesbelow the sources in the −y direction.)

3

As the distance between two sources increases, does the number of fringes increase, decrease, orremain the same?

As the distance between two sources increases, do the fringes get further apart or do they get closertogether?

Experiment 2 – Measuring fringe spacing for double-slit interferenceIf you define a y-axis at the top of the animation window and mark the center of each fringe, then thespacing between fringes is ∆y, as shown in Figure 5. It is this fringe spacing that you will measure in thisexperiment.

Figure 5: Definition of fringe spacing.

When you shine a laser onto 2 slits, you will get an interference pattern for the reasons described above.You will see a pattern like the one shown in Figure 6.

The graph is the intensity of light I as a function of position y on a screen. y = 0 is defined as thecenter of the interference pattern, along a perpendicular bisector of the two slits. Use the graph to answer

4

Figure 6: An interference pattern due to laser light incident on 2-slits.

the following questions.

Note that the center of the left fringe is at y = −10 mm and the center of the right fringe is aty = 10 mm.

What is the distance from the center of the far left fringe to the center of the far right fringe?

How many total fringes are on the screen? (Include the far left and far right fringes.)

What is the slit spacing ∆y between fringes?

The fringe spacing ∆y depends on the wavelength of the light and the distance between the slits (d). Apicture of the apparatus is shown in Figure 7.

5

Figure 7: Experimental apparatus.

If the distance from the slits to the screen is L and the distance between slits is d, then the fringe spacing∆y is

∆y =λL

d

You can solve this for wavelength. In this experiment, you will measure the slit spacing and fringe spacingand will calculate the wavelength.

λ =d∆y

L

Procedure1. Use ring stands to set up a laser and slit film so that light from the laser passes through two slits and

creates an interference pattern on a screen. At first, use the set of slits that are closest together (i.e.smallest slit spacing, d).

2. Tape a piece of paper on a screen that is roughly 2 - 3 m from the slit film (the further the better).You will use this paper to mark bright fringes.

3. Measure L, the distance between the slit film and the paper using a meterstick or measuring tape.

4. On the paper, mark the locations of N fringes where N is enough fringes to span a large distance onthe paper, perhaps about 20 cm or more. The more fringes you use, the less your uncertainty in yourmeasurement of the distance between fringes.

5. Remove the piece of paper and use a ruler to measure the total distance from the first mark to the lastmark.

6. Calculate the spacing between consecutive fringes, ∆y.

7. Calculate the wavelength of the laser, λ. Convert its units to nm.

8. Change the slit spacing d by moving the slit film so that the light passes through a different set of slitsand repeat the measurements.

6

slit spacing, d(mm)

Number offringes (N)

distance be-tween N fringes(cm)

fringe spacing,∆y (cm) L (m) λ (nm)

0.1760.350.701.40

Analysis1. What is the average wavelength of the laser as determined by your measurements using Young’s Double

Slit Experiment?

2. What is the standard deviation of the mean wavelength?

Using a spectrometer to measure wavelengthA spectrometer is a device that uses a diffraction grating to split light into its component wavelengths. Itmeasures the light intensity at different wavelengths. A graph of intensity vs. wavelength is used to identifydifferent wavelengths (i.e. colors) in the spectrum. Note that it only looks at the first order (m = 1)spectrum.

1. An Ocean Optics spectrometer has been set up for you at your computer. It is plugged into thecomputer via the USB port. It has a fiber optic cable attached to it.

2. View Logger Pro. It may already be set up and configured.

3. By default, Logger Pro plots absorbance vs. wavelength. But we wish to measure intensity. Logger Promay already be set up to plot intensity vs. wavelength. If it is not, then go to Experiment→ChangeUnits→Spectrometer: 1→Intensity to change what is being measured by the spectrometer.

4. Point the laser into the end of the fiber optic cable. Use the clothespin to press the button so that thelaser remains on.

7

5. Click the Collect button. After you get a spectrum, click the Stop button. If the graph you see has apeak that goes off the graph (as shown below), that means we’ve saturated the spectrometer, and weneed a way to make the laser less bright.

To do this, we can slightly tilt the laser so that less light enters the spectrometer. Do this and takeanother measurement. Continue trying until you get a peak that is not saturated.

6. Sketch and label the graph you see (when the spectrometer is not saturated).

7. You can measure the wavelength of any point on the spectrum using the Examine tool. Go toAnalyze→Examine. Hover over any point on the graph and the wavelength value will be displayed.Once you hover over a selected peak, you should view the data table to confirm the wavelengths ofeach peak. You can also highlight the peak and click the ‘Statistics’ icon in the toolbar to obtain thewavelength of the maximum intensities.

8. Record the wavelength of the maximum intensity on the data sheet. The fact that the spectrum ofa laser gives a single, narrow peak means that the light from the laser is composed of a single color.Thus it is monochromatic.

Analysis1. What is the wavelength of the laser according to your measurement with the spectrometer?

2. Does your measurement using the double-slit apparatus agree with your measurement using the spec-trometer? In other words, is the measurement with the spectrometer within the uncertainty of yourmeasurement with the double-slit apparatus?

8